Lời giải:Lấy PT$(1)*2$ trừ PT$(2)$ ta có:
$(2\sqrt{5}+4)x+2y-(2y-x)=6-2\sqrt{5}-(6-2\sqrt{5})$
$\Leftrightarrow (5+2\sqrt{5})x=0$
$\Rightarrow x=0$
$\Rightarrow y=3-\sqrt{5}$
Vậy........
Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}x^2+1+y^2+xy=y\\x+y-2=\frac{y}{1+x^2}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x^4+8y^4-2x-y=0\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2+y^2=\frac{1}{5}\\4x^2+3x-\frac{57}{25}=-y\left(3x+1\right)\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\sqrt{12-y}+\sqrt{y\left(12-x\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x-2\sqrt{2}y=\sqrt{5}\\\sqrt{2}x+y=1-\sqrt{10}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}y=\sqrt{3}\end{matrix}\right.\)
giải các hệ phương trình
a)\(\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{4}y-2=0\\5x-y=11\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\sqrt{3x}-\sqrt{5y}=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
Giải hệ pt:
a)\(\left\{{}\begin{matrix}x^2+y^2+x+y=18\\x\left(x+1\right).y\left(y+1\right)=72\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\3y-1=xy\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}2x+3y=xy+5\\\frac{1}{x}+\frac{1}{y+1}=1\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\sqrt{\frac{x}{y}}-3\sqrt{\frac{y}{x}}=2\\x-y+xy=1\end{matrix}\right.\) e)\(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
HELP ME :((
a)\(\left\{{}\begin{matrix}2x+\left|y\right|=3\\x-y=6\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\sqrt{3}x+y=\sqrt{2}\\\sqrt{3}x-\sqrt{2}y=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\sqrt{x+3}+\sqrt{y^2-4y+4}=2\\\sqrt{x+3}-3\left|2-y\right|=1\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}\sqrt{11}x+\sqrt{5}y=16\\3x-2y=-2\sqrt{5}+3\sqrt{11}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{y^2-7x-6}-\sqrt[3]{y\left(x-6\right)}=1\\\sqrt{2\left(x-y\right)^2+6x-2y+4}-\sqrt{y}=\sqrt{x+1}\end{matrix}\right.\)
Giải hệ phương trình
a. \(\left\{{}\begin{matrix}\left(2-\sqrt{3}\right)x-3y=2+5+\sqrt{3}\\4x+y=4-2\sqrt{x}\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}x\sqrt{2}-y\sqrt{3}=1\\x+y\sqrt{3}=2\end{matrix}\right.\)
giải hpt \(\left\{{}\begin{matrix}2\left(x-2\right)^2+\dfrac{1}{\sqrt{y+5}}=3\\\left(x-2\right)^2-\dfrac{2}{\sqrt{y+5}}=-1\end{matrix}\right.\)